Exams Joint Entrance Examination Mathematics Definite Integration Question 160 Question 160 of 246 Medium MCQ
If the real part of the complex number \[{(1 - \cos \theta + 2i\sin \theta )^{ - 1}}\] is \[{1 \over 5}\] for \[\theta \in (0,\pi )\], then the value of the integral \[\int_0^\theta {\sin x} dx\] is equal to:
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\[z = {1 \over {1 - \cos \theta + 2i\sin \theta }}\]
\[ = {{2{{\sin }^2}{\theta \over 2} - 2i\sin \theta } \over {{{(1 - \cos \theta )}^2} + 4{{\sin }^2}\theta }}\]
\[ = {{\sin {\theta \over 2} - 2i\cos {\theta \over 2}} \over {2\sin {\theta \over 2}\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}}\]
\[{\mathop{\rm Re}\nolimits} (z) = {1 \over {2\left( {{{\sin }^2}{\theta \over 2} + 4{{\cos }^2}{\theta \over 2}} \right)}} = {1 \over 5}\]
\[\sin {{2\theta } \over 2} + 4{\cos ^2}{\theta \over 2} = {5 \over 2}\] 1 − cos 2 θ 2 + 4 cos θ 2 = 5 2 1 - {\cos ^2}{\theta \over 2} + 4\cos {\theta \over 2} = {5 \over 2} 1 − cos 2 2 θ + 4 cos 2 θ = 2 5 3 cos 2 θ 2 = 1 2 3{\cos ^2}{\theta \over 2} = {1 \over 2} 3 cos 2 2 θ = 2 1 θ 2 = n π ± π 4 {\theta \over 2} = n\pi \pm {\pi \over 4} 2 θ = nπ ± 4 π θ = 2 n π ± π 2 \theta = 2n\pi \pm {\pi \over 2} θ = 2 nπ ± 2 π \theta \in (0,\pi ) (
\] \therefore θ = π 2 \theta = {\pi \over 2} θ = 2 π ∫ 0 π 2 sin θ d θ = [ − cos θ ] 0 π 2 \int\limits_0^{{\pi \over 2}} {\sin \theta d\theta = {{[ - \cos \theta ]}_0}^{{\pi \over 2}}} 0 ∫ 2 π sin θ d θ = [ − cos θ ] 0 2 π = - (0 - 1) (
\] = 1 Answer the question to unlock